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24 September, 15:46

The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.500 atm, and the small bulb, with a volume of 1.50 L, contains oxygen at a pressure of 2.50 atm. The temperature at the beginning and the end of the experiment is 22 °C. what are the partial gasses of No, and No2?

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  1. 24 September, 19:32
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    P (O2) = 0.300 atm

    P (NO2) = 0.400 atm

    P (NO) = 0 atm

    Explanation:

    Step 1: Data given

    Volume of large bulb = 6.00L

    Large bulb contains nitric oxide at 0.500 atm

    Volume of the small bulb = 1.50 L

    Small bulb contains oxygen at 2/50 atm

    The initial temperature = 22.0°C

    Step 2: The balanced equation

    2 NO + O2 → 2 NO2

    Step 3: Calculate moles of NO

    P*V=n*R*T

    n = (P*V) / (R*T)

    ⇒ with P = the pressure of NO = 0.500 atm

    ⇒ with V = the volume of NO = 6.00L

    ⇒ with R = the gas constant = 0.08206 L*atm/K*mol

    ⇒ with T = the temperature = 22.0 °C = 295 K

    ⇒ with n = the number of moles of NO

    n (NO) = (0.500 * 6.00) / (0.08206*295)

    n (NO) = 0.124 moles

    Step 4: Calculate moles of O2

    n = (P*V) / (R*T)

    ⇒ with P = the pressure of O2 = 2.50 atm

    ⇒ with V = the volume of 02 = 1.50 L

    ⇒ with R = the gas constant = 0.08206 L*atm/K*mol

    ⇒ with T = the temperature = 22.0 °C = 295 K

    ⇒ with n = the number of moles of O2

    n (O2) = (2.50*1.50) / (0.08206*295)

    n (O2) = 0.155 moles

    Step 5: Calculate the limiting reactant

    For 2 moles NO consumed, we need 1 mole O2 to produce 2 moles of NO2

    No is the limiting reactant. It will completely be consumed. (0.124 moles)

    O2 is in excess. There will react 0.124/2 = 0.062 moles O2

    There will remain 0.155 - 0.062 = 0.093 moles O2

    Step 6: Calculate moles of NO2

    For 2 moles NO consumed, we need 1 mole O2 to produce 2 moles of NO2

    For 0.124 moles NO, we'll have 0.124 moles NO2

    Step 7: Calculate partial pressure of O2

    P*V = n*R*T

    P = (n*R*T) / V

    ⇒ n = the number of moles O2 = 0.093 moles O2 remain

    ⇒ R = the gas constant = 0.08206 L*atm/K*mol

    ⇒ T = the temperature = 295 Kelvin

    ⇒ V = the volume = 6.00 + 1.50 = 7.50 L

    P (O2) = 0.300 atm

    Step 8: Calculate partial pressure of NO2

    P = (n*R*T) / V

    ⇒ n = the number of moles NO2 = 0.124 moles

    ⇒ R = the gas constant = 0.08206 L*atm/K*mol

    ⇒ T = the temperature = 295 Kelvin

    ⇒ V = the volume = 6.00 + 1.50 = 7.50 L

    P (NO2) = 0.400 atm

    P (NO) = 0 atm because NO is completely consumed.
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